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Efficiency of a Heat Engine: How to Calculate Heat Transfer in a Monotonic Gas
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[QUOTE="Taylor_1989, post: 6151810, member: 384254"] I have attempted to solve algebraically and seem to have gotten a result which, dose not simplify easily. Here is my current working Step 1 : What I know. $$\Delta E_{int}=Q_in+Wd_{on} \ [1]$$ $$\Delta E_{int}=C_v n\Delta T \ [2]$$ $$\Delta Wd_{on}=-\Delta Wd_{by}=-P\Delta V \ [3]$$ In the PV diagram displayed I have three processes [LIST=1] [*]Isotherm path ##B \rightarrow C## [*]Isobaric path ##C \rightarrow A## [*]Isochoric paths ##A \rightarrow B## [/LIST] Step 2 : Solving Isotherm path $$\Delta E_{int}=0$$ $$0=Q_{in}+Wd_{on}=Q_{in}-Wd_{by}$$ $$Q_{in}=Wd_{by}=Pdv$$ $$P=\frac{nR5T_1}{V}$$ $$Qin=\int^{V_2}_{V_1} \frac{nR5T_1}{V} dv=nR5T_1 ln(\frac{V_2}{V_1})$$ finding ##\frac{V_2}{V_1}## $$P_1V_1=nRT_1$$ $$PV_2=nR5T_1$$ $$\frac{V_2}{V_1}=5$$ therefore $$Q_{in}=nR5ln(5)T_1=Q_H$$ Step 3 : Solving Isobaric path $$\Delta E_{int}=Q_{in}-Wd_{by}$$ $$C_vn\Delta T=Q_{in}-P(\Delta T)$$ $$Q_{in}=C_vn\Delta T + nR\Delta T $$ $$(C_v+R)n\Delta T=Q_{in}$$ $$C_p n\Delta T = Q_{in}$$ $$\Delta T=-4T_1$$ $$Q_{in}=-4C_pnT_1$$ Step 4 : Solving Isobaric path $$\Delta E_{int}=Q_{in}$$ $$C_v n \Delta T=Q_{in}$$ $$4C_v n T_1=Q_{in}$$ Step 5 : Solving for Efficiency $$\epsilon = \frac{W}{Q_in}$$ As system is a complet cycle then ##\Delta E_{int}=0## therefore ##W_T=\sum Q_in## $$\epsilon = \frac{(5ln5)R+4C_v-4C_p}{(5ln5)R+4C_v}$$ $$\frac{(5ln5)R-4R}{(5ln5)R+4C_v}=\frac{(5ln5-4)R}{(5ln5)R+4C_v}$$ And this is were I am stuck, I have I missed a step in my workings? Surley I should be able to simplyfy this into an actual percentage [/QUOTE]
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Efficiency of a Heat Engine: How to Calculate Heat Transfer in a Monotonic Gas
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